Windows' Copilot AI solved this problem nicely by using Vieta's formulas rather than using the Cauchy-Schwarz inequality. Its answer is copied below: To show that (-1 \le c \le \frac{13}{3}) given the conditions (ab + bc + ac = 3) and (a + b + c = 5), let’s start by manipulating these equations. 1. Given Equations: [ ab + bc + ac = 3 ] [ a + b + c = 5 ] 2. Express (a + b) in terms of (c): [ a + b = 5 - c ] 3. Substitute (a + b) into the first equation: [ ab + bc + ac = 3 ] [ ab + c(a + b) = 3 ] [ ab + c(5 - c) = 3 ] [ ab + 5c - c^2 = 3 ] 4. Express (ab) in terms of (c): [ ab = 3 - 5c + c^2 ] 5. Consider the quadratic equation (t^2 - (a + b)t + ab = 0) with roots (a) and (b): [ t^2 - (5 - c)t + (3 - 5c + c^2) = 0 ] 6. For (a) and (b) to be real, the discriminant of this quadratic must be non-negative: [ (5 - c)^2 - 4(3 - 5c + c^2) \ge 0 ] [ 25 - 10c + c^2 - 12 + 20c - 4c^2 \ge 0 ] [ -3c^2 + 10c + 13 \ge 0 ] 7. Solve the quadratic inequality (-3c^2 + 10c + 13 \ge 0): [ 3c^2 - 10c - 13 \le 0 ] 8. Find the roots of the quadratic equation (3c^2 - 10c - 13 = 0): [ c = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot (-13)}}{2 \cdot 3} ] [ c = \frac{10 \pm \sqrt{100 + 156}}{6} ] [ c = \frac{10 \pm \sqrt{256}}{6} ] [ c = \frac{10 \pm 16}{6} ] [ c = \frac{26}{6} \quad \text{or} \quad c = \frac{-6}{6} ] [ c = \frac{13}{3} \quad \text{or} \quad c = -1 ] 9. The quadratic inequality (3c^2 - 10c - 13 \le 0) holds between the roots: [ -1 \le c \le \frac{13}{3} ] Therefore, we have shown that (-1 \le c \le \frac{13}{3}).
